Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What are nice examples of topological spaces $X$ and $Y$ such that $X$ and $Y$ are not homeomorphic but there do exist continuous bijections $f: X \mapsto Y$ and $g: Y \mapsto X$?

share|improve this question
7  
Here are a couple of examples from Omar Antolín-Camarena: sbseminar.wordpress.com/2007/10/30/…. He's on MO, too: mathoverflow.net/users/644/omar-antolin-camarena. Maybe he'll want to elaborate. –  Jonas Meyer Jul 5 '10 at 20:24
5  
As you're looking for a list of examples, this should probably be CW –  Charles Siegel Jul 5 '10 at 22:37

7 Answers 7

up vote 13 down vote accepted

Recycling an old (ca. 1998) sci.math post:

" Anyone know an example of two topological spaces $X$ and $Y$ with continuous bijections $f:X\to Y$ and $g:Y\to X$ such that $f$ and $g$ are not homeomorphisms?

Let $X = Y = Z \times \{0,1\}$ as sets, where $Z$ is the set of integers. We declare that the following subsets of $X$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a topology on $X$.

We declare that the following subsets of $Y$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a toplogy on $Y$.

Define $f:X\to Y$ and $g:Y\to X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$ Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.

This example is due to G. Paseman.

David Radcliffe "

More generally, take a space X with three successively finer topologies T, T' and T''. Form two spaces which have underlying set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' .... and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer topology in one sequence to a rougher topology in the other. You can make them bijective, and show that they are obviously non-homeomorphic for a judicious choice of X, T, T', and T''.

Gerhard "Ask Me About System Design" Paseman, 2010.07.05

share|improve this answer
    
If you want non-compact variations on this, e.g. making the two big spaces normal or connected, then choose X and the T's even more judiciously, and add an extra point or two as needed. And check your work, of course! I leave the compact version(s), or their impossibility, to you. Gerhard "Ask Me About System Design" Paseman, 2010.07.05 –  Gerhard Paseman Jul 5 '10 at 23:19

Here's a continuum analogue of Gerhard Paseman's answer: Let $X$ and $Y$ be topological spaces whose underlying sets are $\mathbb{R}$. As topological spaces, $X$ is the disjoint union of the open interval $(0,\infty)$ with a discrete space whose points are nonpositive reals, while $Y$ is the disjoint union of $(-1,0)$, $(1,\infty)$, and a discrete space whose points form the complement of those intervals. Translation by adding one is a continuous bijection from $X$ to $Y$, and also a continuous bijection from $Y$ to $X$, but the two spaces are not homeomorphic.

share|improve this answer
1  
I like the example, but I do not see a good proof that they are not homeomorphic. Is e.g. (3,5) an open set in both X and Y, or just a subset of an open set? Gerhard "Ask Me About System Design" Paseman, 2010.07.05 –  Gerhard Paseman Jul 6 '10 at 6:02
4  
Look at the one-point compactifications, after noting that both spaces are locally compact Hausdorff. For $X$ we get a disjoint union of a circle and uncountably many points, while for $Y$ we get a disjoint union of a wedge of two circle and uncountably many points. If $X$ and $Y$ were homeomorphic, their one-point compactifications would be as well. –  skupers Jul 6 '10 at 9:13
11  
Look at the number of non-singleton connected components: $X$ has one and $Y$ has 2. This is a topological invariant. –  Henno Brandsma Jul 7 '10 at 11:08

You may find the paper "Bijectively related spaces. I. Manifolds" by P. H. Doyle and J. G. Hocking intersting. They cite some related work, which is also worth checking.

There is a later paper "Unusual and bijectively related manifolds" by J. G. Hocking on this topic, which I have not read.

If you are also interested in spaces satisfying only low separation axioms, you may try my own humble work here and here.

share|improve this answer

I've asked myself this question some time ago and found some counterexamples. After having done this I asked myself what the "smallest" counter example would be. First of all notice that if the set of opens in $X$ and in $Y$ have to have the same cardinality. This show that that if the number of opens in $X$ is finite then $f$ and $g$ both have to be homeomorphisms so the cardinality of the set of opens needs to be at least $\aleph_0$. The answer of Gerhard Paseman shows that using $2^{\aleph_0}$ opens is enough. But one can do better, here is an example that actually has $\aleph_0$ opens:

Let $X$ and $Y$ both have $\mathbb Z$ as underlying set, let the opens of $X$ be $\lbrace \mathbb Z_{\geq i} \mid i \in \mathbb N_{>0} \rbrace \cup \lbrace \emptyset, \mathbb Z \rbrace$ and the set of opens of $Y$ be $\lbrace \mathbb Z_{\geq i} \mid i \in \mathbb N_{>0} , i\neq 2\rbrace \cup \lbrace \emptyset, \mathbb Z \rbrace$, then $f:X\to Y$ given by $f(x)=x$ is continues, and so is $g:Y\to X$ given by $g(a)=a-2$. To see that $X$ and $Y$ are not homeomorphic note that $X$ contains only 1 element that is contained in exactly 2 opens, namely the element $1$, but in $Y$ both the element $1$ and the element $2$ are contained in exactly 2 opens.

share|improve this answer

I don't have my copy of Kelley handy but I think in chapter 1 he gives the example where X is a countable disjoint union of open intervals and a countable discrete set while Y is a countable disjoint union of left-closed, right-open intervals and a countable discrete set.The point is that you can get a half closed interval from an open one by attaching an endpoint and you can build an open interval using a sequence of half closed intervals.

share|improve this answer

Here is an example which comes from using the spaces of Charles Siegel's post.

One have a continuous bijection from [0,1) to the circle given by the exponential function ( t-->exp(2ipit) ). The idea is to use this to construct our spaces. Take A to be a wedge of countably many (one for each integer) [0,1) attached at 0. Let A_n be the same wedge but replacing the [0,1) corresponding to the integers from 1 to n by circles.

X is going to be the disjoint union of A_2, A_4, A_6,... and countably many copies of A. Y is going to be the disjoint union of A_1, A_3, A_5,... and countably many copies of A.

We have a continuous bijection from A_n to A_{n+1} given by replacing the copy of [0,1) corresponding to the integer n+1 by a circle (as it is in A_{n+1}). Using this maps we are going to define f and g.

f is going to map one copy of A to A_1, A_2 to A_3, A_4 to A_5,... and so on, and the rest of the copies of A to the rest of the copies of A in Y. g is going to map A_1 to A_2, A_3 to A_4,... and so on, and the copies of A to the copies of A.

share|improve this answer

My favorite, which is on the wikipedia page for "homeomorphism", is $\phi:[0,2\pi)\to S^1$, by $\phi(\theta)=(\cos\theta,\sin\theta)$, which is continuous and bijective, but not a homeomorphism.

share|improve this answer
6  
The OP also wants a continuous bijection in the other direction. –  Qiaochu Yuan Jul 5 '10 at 22:38
    
Ahh! Missed that. My apologies. –  Charles Siegel Jul 5 '10 at 22:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.